Polar Derivation of an Ellipse

We will show that an ellipse with one focus at the origin can be written in polar coordinates as $$r(\theta)=\frac{\frac{b^{2}}{a}}{1-\frac{c}{a}\cdot cos(\theta)}$$ where $a$ is the semi-major axis, $b$ is the semi-minor axis, $c$ is half of the distance between foci, and $r$ and $\theta$ are the magnitude and angle from the origin to any point on the ellipse.

PEllipse_1.png
Figure 1: Polar derivation of ellipse. Vector $\mathbf{r}$ has length $r$. The distance between focal points is $2c$ and the length of the semi-major axis will be $a$. We know how to compute the length of $\ell$ as the third side of a triangle when the vectors of the two other sides are available. $$\ell=\left\Vert \mathbf{r}-\left(\begin{array}{c} 2c\\ 0 \end{array}\right)\right\Vert =\left\Vert \left(\begin{array}{c} r\cdot \cos(\theta)\\ r\cdot \sin(\theta) \end{array}\right)-\left(\begin{array}{c} 2c\\ 0 \end{array}\right)\right\Vert .$$

Begin with focal points, $F_{1}$ and $F_{2}$ and an arbitrary point, $P$ somewhere on the elliptic curve. For convenience, put $F_{1}$ at the origin and let $a$ be the semi-major axis length. From the definition of the curve write, $$\Vert P-F_{1}\Vert+\Vert P-F_{2}\Vert=2a \tag{1} \label{1}$$ Then we get the graph shown in figure 1. The goal of any polar derivation is to develop an equation in $r$ and $\theta$ and then solve it for $r$.

Since $F_{1}$ is at the origin, vector $\mathbf{r}$ which in the figure terminates at point $P$ is $$\mathbf{r}=\left(\begin{array}{c} r\cdot cos(\theta)\\ r\cdot sin(\theta) \end{array}\right)$$ So equation $\eqref{1}$ can become $$\mathbf{\Vert r}-F_{1}\Vert+\Vert\mathbf{r}-F_{2}\Vert=2a$$ and when we expand the vectors, $$\left\Vert \left(\begin{array}{c} r\cdot cos(\theta)\\ r\cdot sin(\theta) \end{array}\right)-\left(\begin{array}{c} 0\\ 0 \end{array}\right)\right\Vert +\left\Vert \left(\begin{array}{c} r\cdot cos(\theta)\\ r\cdot sin(\theta) \end{array}\right)-\left(\begin{array}{c} 2c\\ 0 \end{array}\right)\right\Vert =2a \tag{2} \label{2}$$ Since the length of vector $\mathbf{r}=r$, \eqref{2} can be written as $$r+\left\Vert \left(\begin{array}{c} r\cdot cos(\theta)\\ r\cdot sin(\theta) \end{array}\right)-\left(\begin{array}{c} 2c\\ 0 \end{array}\right)\right\Vert =2a \tag{3} \label{3}$$ Using algebra we can rearrange $\eqref{3}$ to get $$r-2a=-\left\Vert \left(\begin{array}{c} r\cdot cos(\theta)\\ r\cdot sin(\theta) \end{array}\right)-\left(\begin{array}{c} 2c\\ 0 \end{array}\right)\right\Vert . \tag{4} \label{4}$$ Expanding the right side, $$\begin{aligned} r-2a&=-\sqrt{\left(r\cdot cos(\theta)-2c\right)^{2}+\left(r\cdot sin(\theta)-0\right)^{2}}\\ &=-\sqrt{r^{2}cos^{2}(\theta)-4rc\cdot cos(\theta)+4c^{2}+r^{2}sin^{2}(\theta)}\\ &=-\sqrt{r^{2}-4cr\cdot cos(\theta)+4c^{2}} \end{aligned} \tag{5} \label{5} $$ Then squaring both sides, $$r^{2}-4ar+4a^{2}=r^{2}-4cr\cdot cos(\theta)+4c^{2} \tag{6} \label{6}$$ From \eqref{6} it is simple algebra to solve for $r$ as a function of some constants and $\theta$. $$r(\theta)=\frac{a^{2}-c^{2}}{a-c\cdot cos(\theta)} \tag{7} \label{7}$$

PEllipse2.png
Figure 2: Polar ellipse graph arranged to show a right angle for the semi-minor axis, $b$. In this particular condition, side $\ell=a$, the length of the semi-major axis.

As shown in figure 2, the semi-minor axis, $b$, can be derived from a Pythagorean relation such that $b^{2}=a^{2}-c^{2}$. We actually have that in the numerator of $\eqref{7}$, which allows the use of the semi-minor axis length in the equation. $$r(\theta)=\frac{b^{2}}{a-c\cdot cos(\theta)} \tag{8} \label{8}$$ We can multiply $\eqref{8}$ by $(1/a)$ in both numerator and denominator to get $$r(\theta)=\frac{\frac{b^{2}}{a}}{1-\frac{c}{a}\cdot cos(\theta)} \tag{9} \label{9}$$ the later being done so as to define eccentricity and “parameter p” and perform a substitution. $$eccentricity=\mathrm{e}=\frac{c}{a}$$ $$parameter=p=\frac{b^{2}}{a}$$ $$r(\theta)=\frac{p}{1-\mathrm{e}\cdot\cos(\theta)} \tag{10} \label{10}$$

Aside: The numerator of $\eqref{9}$ is often differently expressed in reference material. It might be bundled into a constant as $$p=\frac{b^{2}}{a}$$ or if $b^{2}$ is replaced by $a^{2}-c^{2}$, and multiplied by $a/a$, $$\frac{b^{2}}{a}=\frac{a}{a}\cdot\frac{a^{2}-c^{2}}{a}=\frac{a(a^{2}-c^{2})}{a^{2}}=a\left(1-\frac{c^{2}}{a^{2}}\right)-a(1-\mathrm{e}^{2}),$$ then it can even be displayed as a function of a and $\mathrm{e}$.